Problem: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its coordinates are $x=3t+2$ and $y=2t^3-2t+4$. What is the particle's acceleration vector at $t=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(0,12)$ (Choice B) B $(0,0) $ (Choice C) C $(2,-2) $ (Choice D) D $(0,12)$
Explanation: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's coordinates are $x=3t+2$ and $y=2t^3-2t+4$, which means its position vector is $(3t+2, 2t^3-2t+4)$. We are asked to find the particle's acceleration vector at $t=0$. In other words, we need to find $\vec{a}(0)$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(3t+2),\dfrac{d}{dt}(2t^3-2t+4)\right) \\\\ &=(3,6t^2-2) \end{aligned}$ Finding $\vec{a}(t)$ $\begin{aligned} \vec{a}(t)&=\dfrac{d}{dt}\vec{v}(t) \\\\ &=\left(\dfrac{d}{dt}(3),\dfrac{d}{dt}(6t^2-2)\right) \\\\ &=(0,12t) \end{aligned}$ Finding $\vec{a}(0)$ $\begin{aligned} \vec{a}({0})&=(0,12({0})) \\\\ &=(0,0) \end{aligned}$ In conclusion, the particle's acceleration vector at $t=0$ is $(0,0)$.